J Algebr Comb (2006) 24:285–297
DOI 10.1007/s10801-006-0019-2
Partial geometries pg(s, t, 2 ) with an abelian Singer
group and a characterization of the van Lint-Schrijver
partial geometry
S. De Winter
Received: 8 September 2005 / Accepted: 13 February 2006 / Published online: 11 July 2006
C Springer Science + Business Media, LLC 2006
Abstract Let S be a proper partial geometry pg(s, t, 2), and let G be an abelian
group of automorphisms of S acting regularly on the points of S. Then either t ≡ 2
(mod s + 1) or S is a pg(5, 5, 2) isomorphic to the partial geometry of van Lint and
Schrijver (Combinatorica 1 (1981), 63–73). This result is a new step towards the
classification of partial geometries with an abelian Singer group and further provides
an interesting characterization of the geometry of van Lint and Schrijver.
Keywords Partial geometry . Abelian Singer group . Geometry of van Lint-Schrijver
1. Introduction and motivation
One of the most important and longstanding conjectures in the theory of finite projective planes is certainly the following:
Any finite projective plane admitting an abelian group acting regularly
on its points must be Desarguesian.
Although with the techniques of today it seems impossible to prove this conjecture,
several weaker versions have been proved and an extensive literature on the subject
exists (we refer to [9] for a recent survey containing extensive bibliographic information).
More generally it is now natural to wonder what happens in the case of
other geometries admitting a(n) (abelian) Singer group. This motivated Ghinelli
[8] to study generalized quadrangles (GQs) admitting a (not necessarily abelian)
The author is Postdoctoral Fellow of the Fund for Scientific Research Flanders (FWO-Vlaanderen).
S. De. Winter ( )
Department of Pure Mathematics and Computer Algebra, Ghent University, Krijgslaan 281-S22,
B-9000 Gent, Belgium
e-mail: sgdwinte@cage.UGent.be
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Singer group. At this point we want to mention that looking at GQs is natural
as both projective planes and GQs are members of the larger class of generalized
polygons.
Recently a question of J. C. Fisher at the 2004 conference “Incidence Geometry”
(May, 2004, La Roche, Belgium) inspired the author and K. Thas to study this subject
in the case of GQs. In [7] they obtained that every finite GQ which admits an abelian
Singer group necessarily arises as the generalized linear representation of a generalized
hyperoval. Further it was noted in that paper that no finite generalized n-gon with n > 4
can admit an abelian Singer group.
In the present paper we will study partial geometries pg(s, t, 2) admitting an abelian
Singer group. This is a natural generalization since partial geometries generalize
projective planes as well as GQs. Further this study is also motivated by the existence of an interesting example: the so-called van Lint-Schrijver partial geometry
[12].
Before proceeding we will now provide some definitions and notation.
A partial geometry pg(s, t, α) is a finite partial linear space S of order (s, t), s, t ≥ 2,
such that
r for every antiflag ( p, L) of S there are exactly α > 0 lines through p intersecting
L.
Partial geometries were introduced in 1963 by Bose [3]. It is easily seen that these geometries have a strongly regular point graph srg(v = (s + 1)(st + α)/α, s(t + 1), s −
1 + t(α − 1), α(t + 1)). Partial geometries for which α = 1 are known as generalized
quadrangles (GQs). A partial geometry will be called proper if 1 < α < min(s, t). If
two distinct points x and y of S are collinear this will be denoted by x ∼ y, while
the line determined by these points will be denoted by x, y. The maximum size of a
clique in the point graph of S equals s + 1, and if C is a clique of this size, then every
point p not belonging to C is adjacent to exactly α elements of C (this is Lemma 2 of
[10]). Finally we mention two constructions of partial geometries.
Let R = {PG(0) (m, q), PG(1) (m, q), . . . , PG(t) (m, q)}, t ≥ 1, be a set of mutually
disjoint PG(m, q) in PG(n, q). We say that R is a PG-regulus if and only if the
following condition is satisfied.
r If PG(m + 1, q) contains PG(i) (m, q), i = 0, 1, . . . , t, then it has a point in common
with α > 0 elements of R \ {PG(i) (m, q)}.
PG-reguli are a special case of the more general class of SPG-reguli which were introduced in 1983 by J. A. Thas [16] (these SPG-reguli give rise to so-called semipartial
geometries). Now suppose that R is a PG-regulus in := PG(n, q) and embed as a
hyperplane in PG(n + 1, q). Define S to be the geometry with as point set the set of all
points of PG(n + 1, q) \ , with as line set the set of all PG(m + 1, q) ⊂ PG(n + 1, q)
that are not contained in and intersect in an element of R, and for which the
incidence relation is the natural one. Then S is a partial geometry pg(q m+1 − 1, t, α)
(see Thas [16]).
To end this section we provide a construction of the above mentioned van LintSchrijver partial geometry [12]. Let β be a primitive element of GF(81) and define
γ := β 16 . Define S to be the geometry with point set the set of elements of GF(81),
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with line set the set of 6-tuples (b, 1 + b, γ + b, γ 2 + b, γ 3 + b, γ 4 + b), b ∈ GF(81),
and with as incidence relation symmetrized containment. Then van Lint and Schrijver
have shown that S is a partial geometry pg(5, 5, 2). It is worthwhile to notice that the
point graph of this partial geometry is a cyclotomic graph.
Finally we note that throughout this paper N will denote the set of natural numbers
(including 0) and that N0 will denote the set of natural numbers without 0.
2. A Benson-type theorem for partial geometries
In this section we will provide an analogue for partial geometries of the theorem of
Benson [2] on automorphisms of generalized quadrangles. The proof of this theorem
is analogous to the proof of Benson’s theorem as given in [14] (Section 1.9, page 23),
so we will only give a sketch of the proof here.
Theorem 2.1. Let S be a partial geometry pg(s, t, α), and let θ be any automorphism
of S. Denote by f the number of fixed points of S under θ and by g the number of
points x of S for which x = x θ ∼ x. Then
(1 + t) f + g ≡ (1 + s)(1 + t)
(mod s + t − α + 1).
Proof: Let P = {x1 , x2 , . . . , xv } be the point set of S. If we denote by A the adjacency
matrix of S, then A2 + (t − s + α + 1)A + (t + 1)(α − s)I = α(t + 1)J , with I the
identity matrix and J the matrix with all entries equal to 1. Consequently A has
eigenvalues s(t + 1), −1 − t and s − α, with respective multiplicities m 0 = 1, m 1
and m 2 , where m 1 and m 2 can easily be computed. Next, if we denote by D the
incidence matrix of S, then M := D D T = A + (t + 1)I , and so M has eigenvalues
(1 + s)(1 + t), 0 and t + 1 + s − α, with respective multiplicities m 0 = 1, m 1 and m 2 .
Now let Q be the matrix with the i j-th entry 1 if xiθ = x j and 0 otherwise. Then Q is
a permutation matrix and one can show that Q M = M Q (see [14]). Consequently, if
θ has order n, then (Q M)n = Q n M n = M n . One deduces that the eigenvalues of Q M
are the eigenvalues of M multiplied with the appropriate roots of unity. From Q M J =
M J = (s + 1)(t + 1)J it follows that (s + 1)(t + 1) will be an eigenvalue of Q M. By
m 0 = 1, this eigenvalue will have multiplicity 1. Further 0 will be an eigenvalue of Q M
with multiplicity m 1 . Let d be a divisor of n. Then the multiplicity of the eigenvalue
ξd (s + t − α + 1) of Q M, will only depend on d and not on the primitive dth rooth of
unity ξd . Denote this multiplicity by ad . Further, as the sum of all dth primitive roots
of unity is an integer Ud [11] we obtain that tr(Q M) = d| n ad (s + t + α − 1)Ud +
(s + 1)(t + 1). On the other hand it is clear that tr(Q M) = (t + 1) f + g. The theorem
follows.
We will apply this theorem to the case of partial geometries admitting a regular
abelian group of automorphisms, but first we shall obtain an easy lemma which will
be used frequently in the rest of this paper without further notice.
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Lemma 2.2. Let S be a partial geometry admitting a regular abelian group of automorphisms G. Suppose g ∈ G. If x g ∼ x for some point x of S, then y g ∼ y for every
y of S.
Proof: By the regularity of G there exists an h ∈ G such that x h = y. As h is an
automorphism of S and as G is abelian, we obtain x h ∼ x gh = x hg = y g , that is,
y ∼ yg.
Corollary 2.3. If S is a partial geometry pg(s, t, α), α = s + 1, and S admits a
regular abelian group G of automorphisms, then
(s + 1)
st + α
≡ (1 + s)(1 + t) ≡ 0
α
(mod s + t − α + 1).
Proof: First notice that if an element of G has a fixed point, then it is necessarily
the identity. To obtain the first equivalence, choose any g ∈ G \ {id} for which there
exists a point x such that x ∼ x g , and apply Theorem 2.1. In order to obtain the
second equivalence, choose a g ∈ G for which there does not exist an x such that
x ∼ x g (notice that g exists as α = s + 1).
3. Subdivision into three classes
The following basic observation will yield a natural subdivision into three classes of
the pairs (S, G), with S a partial geometry and G an abelian automorphism group of
S acting regularly on the points of S.
Lemma 3.1. Let S be a partial geometry pg(s, t, α) and G an abelian Singer group
of S. Let L be any line of S. Then either |StabG (L)| = 1, or |StabG (L)| = s + 1.
Proof: Let x be any point on L and suppose that g ∈ G \ {id} stabilizes L. Let x h ,
h ∈ G, be any point of L \ {x, x g }. Clearly x hg is a point of L, distinct from x h . It
−1
−1
follows that L h = x h , x hg h = x, x g = L. Consequently h stabilizes L and
|StabG (L)| = s + 1.
Based on the above observation we can introduce the following three classes of
partial geometries S with an abelian Singer group G:
r the pair (S, G) is of spread type if |StabG (L)| = s + 1 for each line L of S;
r the pair (S, G) is of rigid type if |StabG (L)| = 1 for each line L of S;
r the pair (S, G) is of mixed type otherwise.
Remark 3.2. Notice that such a division into classes of spread, rigid and mixed type
can be introduced for any partial linear space with an abelian Singer group. (Lemma
3.1 only uses the fact that S is a partial linear space).
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4. Pairs (S, G) of spread-type
In [7] it is shown that for every partial geometry S with α = 1, that is every finite
generalized quadrangle, which admits an abelian Singer group G, the pair (S, G) is of
spread-type, yielding that S is the generalized linear representation of a generalized
hyperoval.
Next suppose that α ≥ 2 and that (S, G) is of spread type. Choose any point x in
S and denote the t + 1 lines through x by L 0 , L 1 , . . . , L t . It readily follows that L iG
determines a spread (of symmetry) of S, explaining the introduced terminology. Let
Si be the stabilizer in G of the line L i , i = 0, 1, . . . , t. Then it is also easily seen that
the pair (G, J = {S0 , S1 , . . . , St }) is an SPG-family in the sense of [6]. From Theorem
2.5 of that same paper [6] it then follows that S is isomorphic to a partial geometry
constructed from a PG-regulus. We will now have a closer look at the parameters in
the case α = 2.
Theorem 4.1. Let S be a partial geometry pg(s, t, 2) and let G be an abelian regular
automorphism group of S. If the pair (S, G) is of spread type, then S is constructed
from a PG-regulus and hence s = q m+1 − 1 for some prime power q and some positive
integer m. Moreover, either t = 2, in which case S is a translation net, or t = 2(s + 2).
In the latter case, q is necessarily a power of 3.
Proof: By Theorem 2.5 of [6] S is constructed from an (S)PG-regulus R, say
an (S)PG-regulus consisting of m-dimensional spaces in PG(n, q), implying
s = q m+1 − 1 for a certain prime power q. Consider two distinct intersecting lines.
The stabilizers in G of these lines have size s + 1 and clearly generate a subgroup
of G of size (s + 1)2 . Consequently v = (s + 1)(st + 2)/2 is divisible by (s + 1)2 .
It follows that (s + 1) divides (st + 2), that is, (s + 1) divides (s + 1)t − t + 2.
Consequently t = z(s + 1) + 2, z ∈ N. If z = 0, then S is a translation net. Next
suppose that z ≥ 1. Let L 0 , . . . , L t denote the t + 1 lines through a fixed point
x. Denote the stabilizers in G of L i by Si , i = 0, 1, . . . , t. The lines L 0 and L 1
determine in a natural way an (s + 1) × (s + 1)-grid L 0 L 1 , which corresponds to
the subgroup S0 S1 of G. As S is a pg(s, t, 2), at most s of the lines L 2 , . . . , L t
can intersect this grid in a point distinct from x. Consequently there exists a
line, say without loss of generality L 2 , intersecting L 0 L 1 only in x. It follows
that S0 S1 S2 is a subgroup of order (s + 1)3 of G. Hence (s + 1)2 should divide
st + 2 = (s + 1)t − t + 2. As we already know that s + 1 divides t − 2, we easily see
that s + 1 must divide (t − 2)/(s + 1) − t. Since t − 2 is divisible by s + 1 we obtain
that s + 1 divides (t − 2)/(s + 1) − 2. If (t − 2)/(s + 1) − 2 = 0, then t = 2(s + 2).
Now assume that s + 1 ≤ (t − 2)/(s + 1) − 2. This implies t ≥ s 2 + 4s + 5. As
on the other hand for partial geometries (s + 1 − 2α)t ≤ (s + 1 − α)2 (s − 1)
(see Chapter 13 of [14]), we see that (s − 3)t ≤ (s − 1)3 . Consequently s = 3
(the case s = 2 yields a dual net, which we do not consider). Substitution
in the identity (s + 1)(t + 1) ≡ 0 (mod s + t − 1) from Corollary 2.3 yields
4(t + 1) ≡ 0 (mod t + 2), and so t + 2 must divide 4. Hence t ∈ {0, 2}, a contradiction. Finally to obtain that q = 3h , it is sufficient to substitute t = 2(s + 2)
in (s + 1)(t + 1) ≡ 0 (mod s + t − 1) from Corollary 2.3. This proves the
theorem.
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Consequently, if S is a proper partial geometry with α = 2 admitting an abelian
Singer group G such that the pair (S, G) is of spread type, then S is a pg(3h(m+1) −
1, 2(3h(m+1) + 1), 2) constructed from a PG-regulus consisting of m-dimensional subspaces of PG(3m + 2, 3h ), for some positive integers m and h (the dimension 3m + 2
of the projective space follows from the definition of a PG-regulus given the parameters
s and t).
Remark 4.2. It is important to notice that there exists a very interesting example of a
PG-regulus R in PG(5, 3) yielding a pg(8, 20, 2). This PG-regulus is due to Mathon
[5] and is the only known PG-regulus yielding a pg(s, 2(s + 2), 2). Further notice that
such a partial geometry has the same parameters as the partial geometry T2∗ (K), which
would arise from a maximal 3-arc K in PG(2, q); it is however well known that such
a maximal arc does not exist, see Cossu [4], Thas [15], Ball, Blokhuis and Mazzocca
[1].
5. Pairs (S, G) of mixed-type
Let S be a proper partial geometry pg(s, t, 2) and G an abelian Singer group of S.
Suppose that the pair (S, G) is of mixed type. Then there are x(s + 1), x ∈ N0 , lines L
through any given point such that the stabilizer of L in G is trivial (as the orbit of any
such line contains exactly s + 1 lines through the given point), and p lines through
any given point with a stabilizer in G of order s + 1. So t + 1 = x(s + 1) + p for
certain x, p ∈ N0 .
Lemma 5.1. Suppose that S is a proper partial geometry pg(s, t, 2) and that G is an
abelian Singer group of S such that the pair (S, G) is of mixed type. Then, with the
above notation, p = 1.
Proof: Suppose by way of contradiction that p = 1. From Corollary 2.3 we obtain
that
x(s + 1)2 + s + 1 ≡ 0
(mod (x + 1)(s + 1) − 2).
Hence, after multiplying with (x + 1)2 we obtain
x [(x + 1)(s + 1)]2 + (x + 1) [(x + 1)(s + 1)] ≡ 0
(mod (x + 1)(s + 1) − 2),
that is,
6x + 2 ≡ 0
(mod (x + 1)(s + 1) − 2).
Consequently 5x + 3 ≥ xs + s, and so s < 5. For s = 3 we find that 6x + 2 must
be divisible by 4x + 2, yielding x = 0, a contradiction. If s = 4, then 6x + 2 has
to be divisible by 5x + 3, from which follows that x = 1 and consequently t = 5.
Substitution in the original identity (s + 1)(t + 1) ≡ 0 (mod s + t − 1) yields 30 ≡
0 (mod 8), the final contradiction.
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Lemma 5.2. If S and G are as in the previous lemma, then p = y(s + 1) + 3 for
certain y ∈ {0, 1}. Further, if y = 1, then also x = 1.
Proof: Consider two distinct intersecting lines with stabilizers in G of size s + 1. As in
Theorem 4.1 we obtain that (s + 1)2 divides v = (s + 1)(st + 2)/2 and consequently
that t = z(s + 1) + 2, that is p = y(s + 1) + 3 for certain y ∈ N. Suppose that y ≥ 1.
Again as in Theorem 4.1 we deduce that (s + 1)3 must divide v, which yields that
t = 2(s + 2), that is x = y = 1.
Remark 5.3. Although the author strongly tends to believe that no partial geometries
pg(s, t, 2) S with an abelian Singer group G exist such that (S, G) is of mixed type,
he has not yet been able to prove this conjecture to be correct. In view of the existence
of pg(s, 2(s + 2), 2), divisibility conditions will clearly not help excluding this case.
6. Pairs (S, G) of rigid type
In this section we will obtain that there is a unique partial geometry pg(s, t, 2) S such
that the pair (S, G) is of rigid type, namely the partial geometry pg(5, 5, 2) of van Lint
and Schrijver.
6.1. The parameters
From now on let S be a proper partial geometry pg(s, t, 2) admitting an abelian
Singer group such that the pair (S, G) is of rigid type. One immediately observes
that t + 1 = x(s + 1) for certain x ∈ N0 (same argument as in the previous section).
Further, Corollary 2.3 implies that (s + 1)(t + 1) ≡ 0 (mod s + t − 1), that is
x(s + 1)2 ≡ 0 (mod (x + 1)(s + 1) − 3).
Multiplying with (x + 1)2 yields
x [(x + 1)(s + 1)]2 ≡ 0
(mod (x + 1)(s + 1) − 3)
from which follows that
9x ≡ 0 (mod (x + 1)(s + 1) − 3).
Hence 9x ≥ (x + 1)(s + 1) − 3 from which we deduce that s < 8.
Theorem 6.1. If S is a proper partial geometry pg(s, t, 2) and G is an abelian Singer
group of S such that the pair (S, G) is of rigid type, then (s, t) = (5, 5).
Proof: We will one by one handle the cases s = 3, 4, . . . , 7.
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r If s = 3 we obtain that 9x must be divisible by 4x + 1 which yields that x = 2 and
hence t = 7. Substituting this in the original identity 2 + st ≡ 0 (mod s + t − 1)
gives us a contradiction.
r In the case where s = 4 we obtain that 9x is divisible by 5x + 2. This implies x = 0,
a contradiction.
r For s = 5 we obtain that 9x is divisible by 6x + 3, and so x = 1 and t = 5.
r Finally, the cases s = 6 and s = 7 can easily be excluded analogously as in the
case s = 4.
6.2. G is elementary abelian
Let S and G be as in the previous subsection, that is, S is a pg(5, 5, 2) and G is
an abelian automorphism group of S acting regularly on the points of S. We will
show that G is elementary abelian. Although we know that s = t = 5, we will in this
subsection always write s, as we believe that this makes general arguments easier to
read.
Now choose any fixed point x in S. We will identify the point y of S with the
unique g ∈ G for which x g = y. Finally, D will denote the set of all points (elements
of G), with exclusion of id, that are collinear with id. We notice that this notation
comes from the theory of partial difference sets, as it is easily checked that D is a
partial difference set in G (see e.g. [13]).
Lemma 6.2. We have D 2 ⊂ D.
Proof: Choose any g ∈ D, and consider the lines L = id, g and L g . The line L g
consists of all points f g with f ∈ L. For any such f with id = f = g the point f g
is collinear with f and g on L. If g 2 were not collinear with id, then necessarily g 2
would be collinear with some f ∈ L \ {id, g}. However this f is already collinear
with g and f g on L g , implying that f would be collinear with more then 2 points on
/ L since otherwise
L g , a contradiction. We conclude that g 2 ∈ D. (Notice that g 2 ∈
the stabilizer of L in G would not be trivial).
Lemma 6.3. Suppose L is a line through id. If we denote the points of L by
id, g1 , g2 , . . . , gs , then L 2 := {id, g12 , g22 , . . . , gs2 } is a set of two by two collinear
points, no three on a line.
Proof: From the previous lemma we know that id ∼ gi2 , for all i ∈ {1, 2, . . . , s}.
We now show that gi2 ∼ g 2j , for i = j. Consider the lines Mi := gi , gi2 and M j :=
g j , g 2j . Then Mi ∩ M j = gi g j . Note that gi g j ∈
/ D since the unique points on L
collinear with gi g j are gi and g j . From the fact that gi ∼ g j it follows that the point
gl gi on Mi is collinear with the point gl g j on M j , for all l ∈
/ {i, j}. Now assume that
gi2 would not be collinear with g 2j . Then gi2 has to be collinear with either g j or gl g j
for some l ∈
/ {i, j}. If gi2 would be collinear with g j , then it would be collinear with
at least three points on L, a contradiction. Would gi2 be collinear with gl g j for some
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l∈
/ {i, j}, then gl g j would be collinear with the distinct points gi g j , gl gi and gi2 on
Mi , again a contradiction. It follows that L 2 is indeed a set of two by two collinear
points. Next, suppose there is a line containing at least three distinct points of L 2 . From
the fact that α = 2 it is immediate that in this case L 2 is a line through id (distinct
−1
from L), say, without loss of generality, L g1 . Hence g2 g1−1 = g 2j for certain j. This
implies that j = 2, as the point g2 g1−1 is collinear with id, g2 and g j on L. But then
g1−1 = g2 , a contradiction. The lemma follows.
Although the following lemma can be avoided, it provides interesting information
on the structure of the geometry S with respect to the group G.
Lemma 6.4. Let the line L be as in the previous lemma. Suppose h ∈ G, with h ∈
/
D ∪ {id}. Then one, and only one, of the following cases occurs:
r h ∈ g 2 , g 2 for unique i and j, i = j;
i
j
r h ∈ gi , g 2 for exactly two values of i.
i
Proof: Suppose h = gi2 , g 2j ∩ gl2 , gk2 or h = gi2 , g 2j ∩ gl , gl2 , where i, j, k, l are
pairwise distinct. In each case gl2 would be collinear with three points on gi2 , g 2j , a
contradiction.
Any point h ∈ gi , gi2 \ {gi , gi2 } can be written as h = g j gi for certain j = i.
Hence, such h also belongs to the line g j , g 2j . Now assume that there would be
a third index k, with i = k = j, such that h ∈ gk , gk2 . Then gk2 would be collinear
with the three distinct points gk gi , h and gi2 on gi , gi2 , a contradiction.
Denote by X 1 the number of elements of G \ (D ∪ {id}) on lines of type gi2 , g 2j and by X 2 the number of elements of G \ (D ∪ {id}) on lines of type gi , gi2 . Easy
counting shows
X 1 = s(s − 1)2 /2
,
X 2 = s(s − 1)/2
and hence
|D ∪ {id}| + X 1 + X 2 = (s + 1)(s 2 + 2)/2 = v.
The lemma now easily follows.
Lemma 6.5. Let the line L be as before. Then gi3 = id for all i.
Proof: First suppose that gi3 does not belong to D ∪ {id}. Since L 2 is a clique of size
s + 1 (and so gi3 is collinear with exactly two points of L 2 ) the previous lemma implies
that either gi3 ∈ gi2 , gk2 , or gi3 = gi , gi2 ∩ gk , gk2 , k = i. In the first case we see that
gk2 ∈ gi2 , gi3 and hence gk2 = g f gi2 for certain f ∈ {1, 2, . . . , s}. Hence gk2 is collinear
with id, gk and g f on L, which implies that f = k. But then gk = gi2 , a contradiction.
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In the second case it follows that gi3 = gi gk , again yielding the contradiction gk = gi2 .
We conclude that gi3 ∈ D ∪ {id}.
Now suppose that gi3 = id. From gi2 ∼ gi3 and gi3 ∈ D it follows that gi3 = gk2 for
certain k = i, or that gi3 = gi . In the first case the point gi3 = gk2 would be collinear
with three distinct points, id, gi and gk on L, a contradiction. The second case is
absurd as well, as it would imply gi2 = id. The lemma is proved.
Corollary 6.6. For all g ∈ G we have g 3 = id
Proof: This follows immediately as each g ∈ G can be written as g = f h, with
f, h ∈ D.
Corollary 6.7. The group G is elementary abelian.
Corollary 6.8. The mapping β : G → G : g → g 2 is an automorphism of G.
Proof: It is clear that β is an endomorphism of G (G is abelian!). We need to show
that β is bijective. Assume by way of contradiction that g 2 = h 2 for certain g = h in
G. By the above we obtain id = g 3 = h 3 = g 2 h. This implies that g 3 = g 2 h, that is
g = h, a contradiction.
Remark 6.9. Although β is an automorphism of G it does not induce an automorphism
of our geometry S. It has nevertheless an interesting geometric interpretation. Let S
be a partial geometry pg(5, 5, 2) admitting an abelian Singer group. It is, using the
above, easily seen that every two distinct collinear points (elements of G) g and h are
contained in two cliques of size 6, namely the one defined by the line L := g, h,
and the one defined by M β , where M = g 2 , h 2 , which is a set of 6 points of S no
three of which are collinear (note that since D 2 = D, g 2 is indeed collinear with h 2 ).
In fact every two distinct collinear points are contained in exactly two distinct cliques
of order 6 as the number of points collinear with two given distinct collinear points
equals 11, and two distinct cliques of size 6 intersect in at most two distinct points.
Define C to be the set of all cliques that arise as the “squares” of lines of S. Then the
geometry with as point set the elements of G, with as line set the elements of C and
with as incidence relation containment is a pg(5, 5, 2) S ∗ and β is an isomorphism
from S to S ∗ .
6.3. The uniqueness of S
Throughout this section S will be a pg(5, 5, 2) and G the elementary abelian group
of order 81 acting regularly on the points of S. We will show that S is unique. We
will again identify the points of S with elements of G. The set D will again be the set
of all elements distinct from id collinear with id. Let L be any line through id. We
denote the points of L by id, g1 , g2 , g3 , g4 and g5 . Further denote by L i , i = 1, . . . , 5
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the line through id containing gi2 . The unique point on L i \ {id} collinear with g j is
the point g j gi2 .
Lemma 6.10. We have g1 g2 g3 g4 g5 = id.
Proof: Consider the point g1 g2 . On L 1 this point is collinear with g12 and g2 g12 and
we have g1 g2 = (g12 )(g2 g12 ). Analogously the two points collinear with g1 g2 on L 2
are uniquely determined (g22 and g1 g22 ). As g1 g2 is already collinear with g12 and g22
it cannot be collinear with gi2 for i ∈
/ {1, 2} (recall that a point exterior to a clique of
order 6 is collinear with exactly 2 points of this clique). Suppose that g1 g2 would be
collinear with g1 g32 . Then, as g1 g2 = (g1 g32 )(g2 g3 ), we would obtain that g2 g3 belongs
to D (recall Lemma 2.2), a contradiction. Hence, using an analogous reasoning, we
obtain that the points collinear with g1 g2 on L 3 , L 4 and L 5 are g4 g32 , g5 g32 , g3 g42 , g5 g42 ,
g3 g52 and g4 g52 . There is a unique h such that g1 g2 = (g4 g32 )h. Moreover h must be
collinear with id (as it maps g4 g32 to a collinear point) and g1 g2 (as g4 g32 belongs to D).
It follows that h must be one of the points collinear with g1 g2 on L 3 , L 4 or L 5 (note that
h cannot belong to L , L 1 or L 2 since g1 g2 = (g1 )(g2 ) = (g12 )(g2 g12 ) = (g22 )(g1 g22 )). We
have that h = g4 g32 since otherwise this would imply that g1 g2 = g3 g42 , a contradiction.
Suppose that h = g5 g32 . Then g1 g2 = g4 g32 g5 g32 = g3 g4 g5 , from which we deduce that
g5 = (g1 g32 )(g2 g42 ). Hence g5 would be collinear with g2 g42 , a contradiction. Further,
as g1 g2 ∈
/ D ∪ {id}, h ∈
/ {g3 g42 , g5 g42 , g3 g52 }. It follows that h = g4 g52 , that is, g1 g2 =
2 2 2
g3 g4 g5 , that is, g1 g2 g3 g4 g5 = id.
Theorem 6.11. The geometry S is unique.
Proof: Supposes S is a pg(5, 5, 2) with a regular abelian automorphism group G.
We need to show that S ∼
= S. For the geometry S we will use the same notation
as above. In the geometry S we will identify the points with the elements of G.
Choose any line L of S through idG and denote its points distinct from idG by gi ,
i = 1, 2, . . . , 5. We define a mapping
G to G as follows: γ (idG ) := idG ,
γ from γ (gi ) := gi , i = 1, 2, . . . , 5, and γ ( i gi ) := i gi . We will show that γ is well
defined and is in fact an isomorphism
between
S and S. In order
to show
that γ is well
defined we should show that i gi = j g j implies that i gi = j g j . It is clear
that the group G is generated by D, hence by {g1 , g2 , g3 , g4 , g5 }, and so, because of
the previous lemma, by {g1 , g2 , g3 , g4 }. Hence every point of S (element of G) can
be expressed in a unique way—as there are exactly 81 points—as g1i1 g2i2 g3i3 g4i4 , with
i k ∈ {0, 1, 2}, k = 1, . . . , 4. From this one easily sees that γ is well defined. And from
the definition of γ it is now also clear that γ is an isomorphism between G and G.
Further, as (with a little abuse of notation) γ (L) = L, L G is the set of all lines of S,
G
and L is the set of all lines of S, is easily deduced that γ is an isomorphism between
S and S.
Corollary 6.12. Let S be a proper partial geometry pg(s, t, 2) and let G be an abelian
group acting regularly on the points of S. If the pair (S, G) is rigid then S is isomorphic
to the partial geometry of van Lint and Schrijver.
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Corollary 6.13. If S is a proper partial geometry pg(s, t, 2) with t ≡ 2 (mod s + 1)
admitting an abelian Singer group, then S is isomorphic to the partial geometry of
van Lint and Schrijver.
Proof: If t ≡ 2
(mod s + 1) the the pair (S, G) must be of rigid type.
Remark 6.14. In the above proof we use the fact that there is a pg(5, 5, 2) with an
abelian Singer group known. This is however not necessary, as our proof is implicitly
constructive. Namely the following can be checked. Let G be the elementary abelian
group of order 81. Let gi , i = 1, . . . , 5 be 5 distinct non-identity elements in G. Suppose that g1i1 g2i2 g3i3 g4i4 g5i5 = id, i k ∈ {0, 1, 2}, if and only if all i k are equal. Finally
define L := {id, g1 , . . . , g5 }. Then the geometry with as point set the elements of
G, as line set the set {L g | g ∈ G} and with as incidence relation containment is a
pg(5, 5, 2). Notice that the existence of such gi is well known. For example consider
the group G ∼
= (Z43 , +) as the set of all 4-tuples over Z3 and define L as the set L :=
{(0, 0, 0, 0), (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (−1, −1, −1, −1)}. This
also gives us an easy geometric interpretation of the van Lint-Schrijver partial geometry: it is the geometry in AG(4, 3) of all translates of the set L.
7. Overview
We can now bring the results of the previous sections together in the following theorem.
Theorem 7.1. Let S be a proper partial geometry pg(s, t, 2), and suppose that there
exists an abelian automorphism group G of S acting regularly on the points of S.
Then either S is a pg(3h(m+1) − 1, 2(3h(m+1) + 1), 2) constructed from a PG-regulus
in PG(3m + 2, 3h ), or S is a pg(s, x(s + 1) + 2, 2) with (S, G) of mixed type, or S is
a pg(5, 5, 2) isomorphic to the partial geometry of van Lint and Schrijver.
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